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Equation of Motion using Graphical Method - Motion Video Lecture - Crash
Text Transcript from Video
now in this lecture we would like to
talk about the equations of the motions
which can be derived from the graph as
well prior to starting with this we
should always keep in mind that the main
thing for this type of equation is the
inferences from the velocity time graph
we should know that the slope of
velocity time graph will always give you
acceleration and the area under the
curve would give it the displacement of
the body within that time now for the
slope to be uniform or acceleration to
be uniform we can say that the line
between VT graph should be a straight
line if the VT graph is not a straight
line then we can confirm that it is not
a uniformly accelerated motion but over
here we are talking of uniformly
accelerated motion for which you have
got the straight line now in this graph
we will say it is a V graph on the y
axis and T X is time mixes the X Exel so
velocity time graph o is the origin use
is the point from where the graph starts
that means the velocity initial velocity
when T equals to 0 is you you know the
initial velocity is termed as u so this
is termed as u ASC is the graph AC is
the line so C comes out to be the final
velocity after time T has passed so C
will be equal to V as we term the final
velocity as V so you can say initial
velocity being u final velocity being V
so AC is the graph now from see we have
dropped the perpendicular which is C B
also a B is a horizontal line so we can
say C B
is basically the difference between the
velocities so we should know also that
the tan theta is given as C B upon a B
perpendicular upon hypotenuse is the tan
beta so taking these things
mathematically we can include that we
have taken that value of a B is equal to
T BC is V minus u and also the tan theta
for this curve will be C B upon a B now
we very well know that acceleration is
given as tan theta which is slope so in
this particular case slope will be equal
to CB upon a B and knowing that CB is V
minus u a B is T so we write this
expression that slope is equal to CB
upon a be V minus u upon T which is
equal to X solution and cross
multiplying we get V is equal to u plus
a T we get the first equation from this
V is equal to u plus 80 or in this
equation you can see that V is the final
velocity U is initial velocity is the
acceleration and T is the time so this
is the final equation of V is equal to u
plus 80 now we talk of the second
equation we very well know whenever a
velocity is varying the average velocity
is defined as the initial velocity plus
final velocity divided by 2 that gives
us the average velocity so average
velocity is defined as V Plus u by T and
the total distance
upon time total is given as average
velocity so we can say the total
displacement will be equal to average
velocity into time this gives us this
value s is equal to V average into T now
we know the value of V average and also
in the expression V is equal to u plus a
T we know V is U plus a T so we place
the values and substituting all the
values over here and giving it
mathematical shape we get the value of
2-u plus 80 divided by 2 into T which on
simplification gives us UT plus half ay
T Square this is the second equation s
is equal to UT plus half KT square also
now we go for the third equation which
is talking of the acceleration initial
velocity final velocity and displacement
since we know that the average velocity
into time gives us the displacement as
we have done in the previous equation
also but from the first equation since
we know that V minus U is 80 or we minus
u upon T will be good acceleration you
place the value of T from that equation
which gives us T is equal to V minus u
upon a we substitute the value of T in
this equation we get the value of time
from this expression and we place the
value of time in this expression so this
is time we place this time in this
equation so we get the further the
equation comes out to be s
is equal to V + u by 2 into V minus u by
2 which is equal to v square minus u
square upon 2 a and cross multiplying we
get this value as v square minus u
square is equal to 2 raised so we can
see that the first equation V is equal
to u plus a T we get directly by taking
the slope average velocity into time
gives as the second equation s is equal
to UT plus half ay T Square and the
third equation we get using v square
minus u square u 2 is by taking the
value of time and using in this
expression now in this lecture we would
like to talk about the equations of the
motions which can be derived from the
graph as well prior to starting with
this we should always keep in mind that
the main thing for this type of equation
is the inferences from the velocity time
graph we should know that the slope of
velocity time graph will always give you
acceleration and the area under the
curve would give it the displacement of
the body within that time now for the
slope to be uniform or acceleration to
be uniform we can say that the line
between V T graph should be a straight
line if the VT graph is not a straight
line then we can confirm that it is not
a uniformly accelerated motion but over
here we are talking of uniformly
accelerated motion for which you have
got the straight line now in this graph
we will say it is a V graph on the y
axis and T X is time mixes the X axial
so velocity time graph o is the origin
use is the point from where the graph
starts that means the velocity
initial velocity when T equals to 0 is
you you know the initial velocity is
termed as u so this is termed as u AC is
the craft AC is the line so C comes out
to be the final velocity after time T
has passed so C will be equal to V as we
term the final velocity as V so you can
say initial velocity being u final
velocity being V so AC is the graph now
from see we have dropped the
perpendicular which is C B also a B is a
horizontal line so we can say C B is
basically the difference between the
velocities so we should know also that
the tan theta is given as C B upon a B
perpendicular upon hypotenuse is the tan
beta so taking these things
mathematically we conclude that we have
taken that value of a B is equal to t BC
is we - you and also the tan theta for
this curve will be C B upon a B now we
very well know that acceleration is
given as tan theta which is slow so in
this particular case slope will be equal
to C B upon a B and knowing that C B is
V minus u a B is C so we write this
expression that slope is equal to C B
upon a be V minus u upon T which is
equal to X solution and cross
multiplying we get V is equal to u plus
80
we get the first equation from this V is
equal to u plus 80 or in this equation
you can see that V is the final velocity
U is the initial velocity is the
acceleration and T is the time so this
is the final equation of V is equal to u
plus 80 now we talk of the second
equation we very well know whenever a
velocity is reading the average velocity
is defined as the initial velocity plus
final velocity divided by two that gives
us the average velocity so average
velocity is defined as V Plus u by T and
the total distance upon time total is
given as average velocity so we can say
the total displacement will be equal to
average velocity into time this gives us
this value today s is equal to V average
into T now we know the value of V
average and also in the expression V is
equal to u plus a T we know V is U plus
a T so we place the values and
substituting all the values over here
and giving it mathematical shape we get
the value of 2-u plus 80 divided by 2
into T which on simplification gives us
UT plus half ay T Square this is the
second equation s is equal to UT plus
half KT square also now we go for the
third equation which is talking of the
acceleration initial velocity final
velocity and displacement since we know
that the average velocity into time
gives us the displacement as we have
done in the previous equation also but
from the first equation since we know
that V minus U is a T or V minus u upon
T will be good acceleration we place the
value of T
from that equation which gives us T is
equal to V minus u upon a we substitute
the value of T in this equation we get
the value of time from this expression
and we place the value of time in this
expression so this is time we place this
time in this equation so we get the
further the equation comes out to be s
is equal to V Plus u by 2 into V minus u
by 2 which is equal to v square minus u
square upon 2 a and cross multiplying we
get this value as v square minus u
square is equal to 2 raised so we can
see that the first equation V is equal
to u plus a T we get directly by taking
the slope average velocity into time
gives as the second equation s is equal
to UT plus half ay T Square and the
third equation we get using v square
minus u square value q es by taking the
value of time and using in this
expression
talk about the equations of the motions
which can be derived from the graph as
well prior to starting with this we
should always keep in mind that the main
thing for this type of equation is the
inferences from the velocity time graph
we should know that the slope of
velocity time graph will always give you
acceleration and the area under the
curve would give it the displacement of
the body within that time now for the
slope to be uniform or acceleration to
be uniform we can say that the line
between VT graph should be a straight
line if the VT graph is not a straight
line then we can confirm that it is not
a uniformly accelerated motion but over
here we are talking of uniformly
accelerated motion for which you have
got the straight line now in this graph
we will say it is a V graph on the y
axis and T X is time mixes the X Exel so
velocity time graph o is the origin use
is the point from where the graph starts
that means the velocity initial velocity
when T equals to 0 is you you know the
initial velocity is termed as u so this
is termed as u ASC is the graph AC is
the line so C comes out to be the final
velocity after time T has passed so C
will be equal to V as we term the final
velocity as V so you can say initial
velocity being u final velocity being V
so AC is the graph now from see we have
dropped the perpendicular which is C B
also a B is a horizontal line so we can
say C B
is basically the difference between the
velocities so we should know also that
the tan theta is given as C B upon a B
perpendicular upon hypotenuse is the tan
beta so taking these things
mathematically we can include that we
have taken that value of a B is equal to
T BC is V minus u and also the tan theta
for this curve will be C B upon a B now
we very well know that acceleration is
given as tan theta which is slope so in
this particular case slope will be equal
to CB upon a B and knowing that CB is V
minus u a B is T so we write this
expression that slope is equal to CB
upon a be V minus u upon T which is
equal to X solution and cross
multiplying we get V is equal to u plus
a T we get the first equation from this
V is equal to u plus 80 or in this
equation you can see that V is the final
velocity U is initial velocity is the
acceleration and T is the time so this
is the final equation of V is equal to u
plus 80 now we talk of the second
equation we very well know whenever a
velocity is varying the average velocity
is defined as the initial velocity plus
final velocity divided by 2 that gives
us the average velocity so average
velocity is defined as V Plus u by T and
the total distance
upon time total is given as average
velocity so we can say the total
displacement will be equal to average
velocity into time this gives us this
value s is equal to V average into T now
we know the value of V average and also
in the expression V is equal to u plus a
T we know V is U plus a T so we place
the values and substituting all the
values over here and giving it
mathematical shape we get the value of
2-u plus 80 divided by 2 into T which on
simplification gives us UT plus half ay
T Square this is the second equation s
is equal to UT plus half KT square also
now we go for the third equation which
is talking of the acceleration initial
velocity final velocity and displacement
since we know that the average velocity
into time gives us the displacement as
we have done in the previous equation
also but from the first equation since
we know that V minus U is 80 or we minus
u upon T will be good acceleration you
place the value of T from that equation
which gives us T is equal to V minus u
upon a we substitute the value of T in
this equation we get the value of time
from this expression and we place the
value of time in this expression so this
is time we place this time in this
equation so we get the further the
equation comes out to be s
is equal to V + u by 2 into V minus u by
2 which is equal to v square minus u
square upon 2 a and cross multiplying we
get this value as v square minus u
square is equal to 2 raised so we can
see that the first equation V is equal
to u plus a T we get directly by taking
the slope average velocity into time
gives as the second equation s is equal
to UT plus half ay T Square and the
third equation we get using v square
minus u square u 2 is by taking the
value of time and using in this
expression now in this lecture we would
like to talk about the equations of the
motions which can be derived from the
graph as well prior to starting with
this we should always keep in mind that
the main thing for this type of equation
is the inferences from the velocity time
graph we should know that the slope of
velocity time graph will always give you
acceleration and the area under the
curve would give it the displacement of
the body within that time now for the
slope to be uniform or acceleration to
be uniform we can say that the line
between V T graph should be a straight
line if the VT graph is not a straight
line then we can confirm that it is not
a uniformly accelerated motion but over
here we are talking of uniformly
accelerated motion for which you have
got the straight line now in this graph
we will say it is a V graph on the y
axis and T X is time mixes the X axial
so velocity time graph o is the origin
use is the point from where the graph
starts that means the velocity
initial velocity when T equals to 0 is
you you know the initial velocity is
termed as u so this is termed as u AC is
the craft AC is the line so C comes out
to be the final velocity after time T
has passed so C will be equal to V as we
term the final velocity as V so you can
say initial velocity being u final
velocity being V so AC is the graph now
from see we have dropped the
perpendicular which is C B also a B is a
horizontal line so we can say C B is
basically the difference between the
velocities so we should know also that
the tan theta is given as C B upon a B
perpendicular upon hypotenuse is the tan
beta so taking these things
mathematically we conclude that we have
taken that value of a B is equal to t BC
is we - you and also the tan theta for
this curve will be C B upon a B now we
very well know that acceleration is
given as tan theta which is slow so in
this particular case slope will be equal
to C B upon a B and knowing that C B is
V minus u a B is C so we write this
expression that slope is equal to C B
upon a be V minus u upon T which is
equal to X solution and cross
multiplying we get V is equal to u plus
80
we get the first equation from this V is
equal to u plus 80 or in this equation
you can see that V is the final velocity
U is the initial velocity is the
acceleration and T is the time so this
is the final equation of V is equal to u
plus 80 now we talk of the second
equation we very well know whenever a
velocity is reading the average velocity
is defined as the initial velocity plus
final velocity divided by two that gives
us the average velocity so average
velocity is defined as V Plus u by T and
the total distance upon time total is
given as average velocity so we can say
the total displacement will be equal to
average velocity into time this gives us
this value today s is equal to V average
into T now we know the value of V
average and also in the expression V is
equal to u plus a T we know V is U plus
a T so we place the values and
substituting all the values over here
and giving it mathematical shape we get
the value of 2-u plus 80 divided by 2
into T which on simplification gives us
UT plus half ay T Square this is the
second equation s is equal to UT plus
half KT square also now we go for the
third equation which is talking of the
acceleration initial velocity final
velocity and displacement since we know
that the average velocity into time
gives us the displacement as we have
done in the previous equation also but
from the first equation since we know
that V minus U is a T or V minus u upon
T will be good acceleration we place the
value of T
from that equation which gives us T is
equal to V minus u upon a we substitute
the value of T in this equation we get
the value of time from this expression
and we place the value of time in this
expression so this is time we place this
time in this equation so we get the
further the equation comes out to be s
is equal to V Plus u by 2 into V minus u
by 2 which is equal to v square minus u
square upon 2 a and cross multiplying we
get this value as v square minus u
square is equal to 2 raised so we can
see that the first equation V is equal
to u plus a T we get directly by taking
the slope average velocity into time
gives as the second equation s is equal
to UT plus half ay T Square and the
third equation we get using v square
minus u square value q es by taking the
value of time and using in this
expression
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Apr 22, 2026 Last updated
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